1. **State the problem:** Given the equation $$8 + \csc^2 \theta = 6 \cot \theta,$$ find the value of $$\tan \theta$$. 2. **Recall trigonometric identities:** - $$\csc^2 \theta = 1 + \cot^2 \theta$$ - $$\tan \theta = \frac{1}{\cot \theta}$$ 3. **Substitute $$\csc^2 \theta$$ in the equation:** $$8 + 1 + \cot^2 \theta = 6 \cot \theta$$ which simplifies to $$9 + \cot^2 \theta = 6 \cot \theta$$ 4. **Rearrange the equation:** $$\cot^2 \theta - 6 \cot \theta + 9 = 0$$ 5. **Solve the quadratic equation:** Let $$x = \cot \theta$$, then $$x^2 - 6x + 9 = 0$$ 6. **Factorize or use quadratic formula:** $$ (x - 3)^2 = 0 $$ So, $$x = 3$$ 7. **Find $$\tan \theta$$:** Since $$x = \cot \theta = 3$$, $$\tan \theta = \frac{1}{3}$$ 8. **Find $$\tan(\theta + 45^\circ)$$ using the formula:** $$\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$$ Here, $$a = \theta$$ and $$b = 45^\circ$$, and $$\tan 45^\circ = 1$$. 9. **Substitute values:** $$\tan(\theta + 45^\circ) = \frac{\frac{1}{3} + 1}{1 - \frac{1}{3} \times 1} = \frac{\frac{1}{3} + \frac{3}{3}}{1 - \frac{1}{3}} = \frac{\frac{4}{3}}{\frac{2}{3}} = 2$$ **Final answers:** - $$\tan \theta = \frac{1}{3}$$ - $$\tan(\theta + 45^\circ) = 2$$